# Will Cadmium hydroxide precipitate from 0.01 M solution of "CdCl"_2 at pH 9. K_(sp) of "Cd"("OH")_2 = 2.5 xx 10^(-14)"M"^3? Please solve,thanks!

Mar 23, 2016

Yes, a precipitate will form.

#### Explanation:

Your strategy here will be to calculate the concentrations of the cadmium(II) cations, ${\text{Cd}}^{2 +}$, and of the hydroxide anions, ${\text{OH}}^{-}$, and figure out if these concentrations are high enough to ensure that a precipitate is formed.

So, you know that the pH of the solution is equal to $9$. As you know, an aqueous solution at room temperature has

$\textcolor{p u r p \le}{| \overline{\underline{\textcolor{w h i t e}{\frac{a}{a}} \textcolor{b l a c k}{\text{pH " + " POH} = 14} \textcolor{w h i t e}{\frac{a}{a}} |}}}$

This means that the pOH of the solution, which can be used to find the concentration of hydroxide anions, will be equal to

$\text{pOH} = 14 - 9 = 5$

This means that you have

["OH"^(-)] = 10^(-"pOH") = 10^(-5)"M"

Cadmium chloride, ${\text{CdCl}}_{2}$, is a soluble ionic compound that dissociates completely in aqueous solution to form cadium(II) cations and chloride anions

${\text{CdCl"_text(2(aq]) -> "Cd"_text((aq])^(2+) |+ 2"Cl}}_{\textrm{\left(a q\right]}}^{-}$

Notice that one mole of cadmium chloride produces one mole of cadmium cations in solution. This means that you have

["Cd"^(2+)] = ["CdCl"_2] = 10^(-2)"M"

The partial dissociation of cadmium hydroxide in aqueous solution looks like this

${\text{Cd"("OH")_text(2(s]) rightleftharpoons "Cd"_text((aq])^(2+) + color(red)(2)"OH}}_{\textrm{\left(a q\right]}}^{-}$

By definition, the solubility product constant, ${K}_{s p}$, for cadmium hydroxide will be equal to

${K}_{s p} = {\left[{\text{Cd"^(2+)] * ["OH}}^{-}\right]}^{\textcolor{red}{2}}$

Now, in order to determine if a precipitate if formed or not, you need to calculate the reaction quotient, ${Q}_{s p}$.

The difference between ${K}_{s p}$ and ${Q}_{s p}$ lies in the fact that the former is calculated using equilibrium concentrations, while the latter is calculated using the concentrations of the ions at a given time.

In other words, ${Q}_{s p}$ can be used to determine if a precipitate will form. In order for a precipitate to form, you need to have

${Q}_{s p} > {K}_{s p}$

At this point, a precipitate will form until you reach saturation, i.e until ${Q}_{s p} = {K}_{s p}$.

If ${Q}_{s p} < {K}_{s p}$, the solution is unsaturated and a precipitate will not form.

Plug in the concentrations of the two ions to get

${Q}_{s p} = {10}^{- 2} {\text{M" * (10^(-5))^color(red)(2)"M"^color(red)(2) = 10^(-12)"M}}^{3}$

Since you have

${10}^{- 12} {\text{M"^3 > 2.5 * 10^(-14)"M}}^{3}$

cadmium hydroxide will indeed precipitate out of solution.